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Average velocity practicing

  1. Aug 6, 2014 #1

    bobie

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    1. The problem statement, all variables and given/known data

    I am just practicing and I found this problem:

    A Body is moving with constant speed v along a circle of radius R. Find the average velocity of the body from time t=0 to t = R/3V.

    Before I try to solve it, can you explain what it means?
    If speed v is constant, isn't average v always just = v?

    Thanks
    btw: the answer is 3/pi v
     
    Last edited: Aug 6, 2014
  2. jcsd
  3. Aug 6, 2014 #2
    Remember that velocity is different from speed. Even though a particle may be traveeling at a constant speed around a circle, its velocity is constantly changing - because its always changing direction.

    Furthermore, the magnitutde of average velocity only equals average speed when the displacement is equal to the distance travelled.
     
    Last edited: Aug 6, 2014
  4. Aug 6, 2014 #3

    bobie

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    That is really interesting, matinee, but, please, tell me:

    -v is constantly changing + or -? isn't the change of v always the same as curvature (the change of direction) is always the same?

    - if the body travels for a radians, isn't the displacement = r? if not, how do you calculate it?

    Does this all mean that the average velocity is different if t is t0 to t=R/3v or twice as much?
     
  5. Aug 6, 2014 #4

    Nathanael

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    Displacement has nothing to do with the path taken. Displacement is simply "the difference between the final and initial positions"

    So if the body travels one radian? The displacement will be a bit less than r
    (it will be the length of the straight or "secant" line, not the length of the arc)


    "Average velocity" (defined over a specific time interval) is just the total displacement divided by the total length of time.

    For example, if you went around pi radians, in one second, your average velocity would 2r m/s (if r is measured in meters)



    P.S.
    Are you asking if the velocity is decreasing or increasing? The answer is neither, the magnitude is constant. The speed is not what is changing, the direction is the "part" that is constantly changing.
     
  6. Aug 6, 2014 #5
    If you know the radius, and the angle [itex] \theta [/itex] then you can calculate x with [itex] x = 2r\sin \frac{\theta}{2} [/itex]. So if a particle travels an angle theta along the arc, then its displacement is x.
     

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  7. Aug 6, 2014 #6

    bobie

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    Does that mean that if I travel 2pi average v is 0? that is cool!
    Now in this example, the result is 3/pi , that means that v is not relevant and the average is .95493.*2 =1.9099
    Let's say that R = 6m and v =2, time is 6/3*2 = 1 sec , right?
    In 1 second B travels 2m, which is 1/3 R , radians : 19.1°
    distance d is 2*sin 9°.549 = .33179 *6= 1.99075 /v = 0.99537
    That is not 3/pi
     
    Last edited: Aug 6, 2014
  8. Aug 6, 2014 #7

    Nathanael

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    Exactly!

    6/(2*3) = 1
    but
    6/3*2 = 4
    (parenthesis are important)

    Well let's not use a specific value for r (you were using 6 for that calculation right?) let's just call it "r"

    So you get 0.33179*r is the displacement. But that's only the displacement, not the average speed.
     
  9. Aug 6, 2014 #8

    bobie

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    I have done d/v .33*6/2 = 1.99075 that's the average
    but it should be 1.99099
    Do you think the result given by OP 3/pi may be wrong?
     
  10. Aug 6, 2014 #9

    Nathanael

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    Oh, sorry about that. Yes I think it is.
     
  11. Aug 7, 2014 #10

    bobie

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    Can you guys tell me if this concept of average v in circular motion is of any practical use.
    Can you mention one instance when it comes in handy?

    Thanks
     
  12. Aug 7, 2014 #11

    ehild

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    I think the only practical use is to make it as a problem in tests, to confuse students.:tongue2:

    ehild
     
  13. Aug 7, 2014 #12
    I totally agree.

    This is one of those types of questions that you may find on a quiz or exam to show them that you truely understand the concept, where you need to be confident in your work and not start second guessing yourself.
     
  14. Aug 7, 2014 #13

    bobie

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    Thanks, a sigh of relief!
     
  15. Aug 7, 2014 #14

    HallsofIvy

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    Perhaps to see if you really understand the concept of "average speed". If you are going around in circles you are not going anywhere so of course your average speed is 0.
     
  16. Aug 7, 2014 #15

    ehild

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    You mixed speed and velocity I am afraid. It is the average velocity that is zero if you go a full circle.

    ehild
     
  17. Aug 7, 2014 #16

    olivermsun

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    Actually, the concept of a vector average of "something" in an oscillatory system is used everywhere in math, physics, engineering, … :wink:
     
  18. Aug 7, 2014 #17

    ehild

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    As for velocity, it is usually in the context that the average is zero. :smile:


    ehild
     
  19. Aug 9, 2014 #18

    bobie

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    Probably this concept is useful to establish that a ball rotating on a string has p = 0 (m * v = 0) ?
     
  20. Aug 9, 2014 #19

    ehild

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    The ball has velocity at every instant so its momentum is not zero. But the time average for a full period is zero.

    ehild
     
  21. Aug 9, 2014 #20

    olivermsun

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    Yes, or close to it. :smile:

    As you know, we're often interested in things that don't quite make it to zero. For example: average the velocity of a certain bit of water (or suspended particle) in a surface wave on the ocean.
     
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